3 Answers
3

You came close. When you have the $3$ chairs to put in $6$ places it becomes as stars and bars problem, so the answer is ${6+3-1\choose6-1}={8\choose5}=56$, which does result in $6720.$

Your solution us incorrect, because ${6\choose3}$ is the number of ways to pick three of the spots in which to put a chair. This would be OK if it were required that there be no more than two chairs between any two of the kids, but nothing prevents us from putting all three chairs between the second at third kid, for example.

We have six spots where we can put chairs, and three chairs to distribute. This is the same problem as distributing three indistinguishable balls in six distinct bins, which is solved by stars and bars.

To answer the point you raise in your last comment, the spots are distinguishable because it makes a difference if Jane and Mary are separated by two chairs or three or four, but it doesn't matter which particular chairs we put between them; the red chair and the green chair are the same as far as we are concerned in this problem.

$\begingroup$Can you explain a little bit more (the stars and bars problem part)? I kinda get what you're trying to say, but I don't understand how this applies to this problem.$\endgroup$
– MoriaAug 10 at 16:14

$\begingroup$Do you understand stars and bars in general?$\endgroup$
– saulspatzAug 10 at 16:15

$\begingroup$yes. But I don't get how you got that the $6$ places are the bars and the $3$ chairs are the stars (indistinguishable). And not the other way around.$\endgroup$
– MoriaAug 10 at 16:21

Once again it is a case of multinomial .
Let x1 be number of chairs before 1 child, $x2$ be number of chairs between 1st and 2nd child , $x3 $between 2nd and 3rd, $x4$ between 3rd
and 4th ,$x5 $between 4th and 5th, $x6 $ after 5th.